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2001 AIME II Problems/Problem 14

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Problem

There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ .

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

Solution

Solution 1

To satisfy $z^{28} - z^{8} - 1 = 0$ , $\text{Im}\,(z^{28})=\text{Im}\,(z^{8})$ and $\text{Re}\,(z^{28})=\text{Re}\,(z^{8})+1$ .

Since $\mid z \mid = 1$ , $z$ is on the unit circle centered at the origin in the complex plane.

Since $\text{Im}\,(z^{28})=\text{Im}\,(z^{8})$ , $z^{28}$ and $z^8$ have the same $y$ coordinate. Since $\text{Re}\,(z^{28})=\text{Re}\,(z^{8})+1$ , $z^{28}$ is $1$ unit to the right of $z^{8}$ . It is easy to see that the only possibilities are $(z^{28},z^{8})=(\text{cis}\,(60),\text{cis}\,(120))$ or $(\text{cis}\,{(300)},\text{cis}\,{(240)})$ .

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For the first possibility:

$\begin{align*}z^{28}=\text{cis}\,(28\theta)=\text{cis}\,(60) \Rightarrow 28\theta \equiv 60 \pmod{360} &\Rightarrow \thet...$

Thus, $\theta \equiv 15 \pmod{90}$ . This yields $\theta = 15, 105, 195, 285$ .

For the second possibility:

$\begin{align*}z^{28}=\text{cis}\,(28\theta)=\text{cis}\,(300) \Rightarrow 28\theta \equiv 300 \pmod{360} &\Rightarrow \th...$

Thus, $\theta \equiv 75 \pmod{90}$ . This yields $\theta = 75, 165, 255, 345$ .

Therefore $(\theta_1,\theta_2,\theta_3,\theta_4,\theta_5,\theta_6,\theta_7,\theta_8)=(15,75,105,165,195,255,285,345)$ and $\theta_2+\theta_4+\theta_6+\theta_8=\boxed{840}$ .

Solution 2

Rearrange the given equation as $z^8\left(z^{20}-1\right) = 1$ ; the magnitudes of both sides must be equal, so

$\left|z^8\left(z^{20}-1\right)\right| = \left|z^{20}-1\right| = \left| 1 \right| = 1$

Thus the distance between $z^{20} = \text{cis}\, 20\theta$ and $(1,0)$ on the coordinate plane is $1$ . By the distance formula,

$1 = \sqrt{(\cos 20\theta - 1)^2 + \sin ^2 20\theta} = \sqrt{2 - 2 \cos 20\theta} \Longrightarrow \cos 20\theta = \frac 12$

And $20\theta = 60, 300 + 360n$ , while $z^{20} - 1 = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i - 1 = \text{cis}\,(120,240)$ . Thus $z^8 = \frac{1}{z^{20}-1} = \text{cis}^{-1}\, \{120,240\} = \text{cis}\,\{240,120\}$ . We thus have $20\theta = 60 + 360n$ and $8\theta = 240 + 360n$ or $20\theta = 300 + 360n$ and $8\theta = 120 + 360n$ . From here, follow the above solution.

See also

2001 AIME II (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2001_AIME_II_Problems/Problem_14"

Category: Intermediate Algebra Problems

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